x^3+3x^2+3x=0 x^3=4x x^3-49x=0

Đáp án:

 $x^3+3x^2+3x=0$

$⇔x(x^2+3x+3)=0$

$⇔x^2+3x+3=0$

$⇔x^2 +2 .x \dfrac{3}{2} + \dfrac{9}{4} + \dfrac{3}{4}=0$

$⇔(x+\dfrac{3}{2})^2 +\dfrac{3}{4} > 0 ∀ x$ (loại)

$⇔x=0$

Vậy $x=0$

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$x^3=4x$

$⇔x^3-4x=0$

$⇔x^3-2x^2+2x^2-4x=0$

$⇔x^2(x-2)+2x(x-2)=0$

$⇔(x-2)(x^2+2x)=0$

$⇔x(x+2)(x-2)=0$

⇔\(\left[ \begin{array}{l}x=0\\x+2=0\\x-2=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=0\\x=-2\\x=2\end{array} \right.\) 

Vậy $\text{x ∈ {0 ; 2 ; -2} }$

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$x^3-49x=0$

$⇔x^3-7x^2+7x^2-49x=0$

$⇔x^2(x-7)+7x(x-7)=0$

$⇔(x-7)(x^2+7x)=0$

$⇔x(x+7)(x-7)=0$

⇔\(\left[ \begin{array}{l}x=0\\x+7=0\\x-7=0\end{array} \right.\) 

⇔\(\left[ \begin{array}{l}x=0\\x=-7\\x=7\end{array} \right.\) 

Vậy $\text{ x ∈ {0 ; -7 ; 7 } }$